Medium MCQ +4 / -1 PYQ · JEE Mains 2023

Identify the correct order of standard enthalpy of formation of sodium halides :

  1. A $\mathrm{NaI}<\mathrm{NaBr}<\mathrm{NaCl}<\mathrm{NaF}$ Correct answer
  2. B $\mathrm{NaI}<\mathrm{NaBr}<\mathrm{NaF}<\mathrm{NaCl}$
  3. C $\mathrm{NaF}<\mathrm{NaCl}<\mathrm{NaBr}<\mathrm{NaI}$
  4. D $\mathrm{NaCl}<\mathrm{NaF}<\mathrm{NaBr}<\mathrm{NaI}$

Solution

Lattice energy is directly proportional to the charges of the ions and inversely proportional to the sum of their radii: <br/><br/> Lattice energy ∝ (Q₁Q₂) / (r₁ + r₂) <br/><br/> As we move down the group, the size of the halide ions increases, resulting in a decrease in lattice energy. This leads to a decrease in the exothermicity of the lattice energy release during the formation of sodium halides. <br/><br/> Thus, the correct order of standard enthalpy of formation of sodium halides, considering both lattice energy and electron affinity, is: <br/><br/> $\mathrm{NaF}>\mathrm{NaCl}>\mathrm{NaBr}>\mathrm{NaI}$

About this question

Subject: Chemistry · Chapter: s-Block Elements · Topic: Alkali Metals

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